I had to write a program that will receive an int 'n' and another one 'd' - and will print the number n with commas every d digits from right to left.
If 'n' or 'd' are negative - the program will print 'n' as is.
I although had to make sure that there is no commas before or after the number and I'm not allowed to use String or Arrays.
for example: n = 12345678
d=1: 1,2,3,4,5,6,7,8
d=3: 12,345,678
I've written the following code:
public static void printWithComma(int n, int d) {
if (n < 0 || d <= 0) {
System.out.println(n);
} else {
int reversedN = reverseNum(n), copyOfrereversedN = reversedN, counter = numberLength(n);
while (reversedN > 0) {
System.out.print(reversedN % 10);
reversedN /= 10;
counter--;
if (counter % d == 0 && reversedN != 0) {
System.out.print(",");
}
}
/*
* In a case which the received number will end with zeros, the reverse method
* will return the number without them. In that case the length of the reversed
* number and the length of the original number will be different - so this
* while loop will end the zero'z at the right place with the commas at the
* right place
*/
while (numberLength(copyOfrereversedN) != numberLength(n)) {
if (counter % d == 0) {
System.out.print(",");
}
System.out.print(0);
counter--;
copyOfrereversedN *= 10;
}
}
}
that uses a reversNum function:
// The method receives a number n and return his reversed number(if the number
// ends with zero's - the method will return the number without them)
public static int reverseNum(int n) {
if (n < 9) {
return n;
}
int reversedNum = 0;
while (n > 0) {
reversedNum += (n % 10);
reversedNum *= 10;
n /= 10;
}
return (reversedNum / 10);
}
and numberLength method:
// The method receives a number and return his length ( 0 is considered as "0"
// length)
public static int numberLength(int n) {
int counter = 0;
while (n > 0) {
n /= 10;
counter++;
}
return counter;
}
I've been told that the code doesn't work for every case, and i am unable to think about such case (the person who told me that won't tell me).
Thank you for reading!
You solved looping through the digits by reversing the number, so a simple division by ten can be done to receive all digits in order.
The comma position is calculated from the right.
public static void printWithComma(int n, int d) {
if (n < 0) {
System.out.print('-');
n = -n;
}
if (n == 0) {
System.out.print('0');
return;
}
int length = numberLength(n);
int reversed = reverseNum(n);
for (int i = 0; i < length; ++i) {
int nextDigit = reversed % 10;
System.out.print(nextDigit);
reversed /= 10;
int fromRight = length - 1 - i;
if (fromRight != 0 && fromRight % d == 0) {
System.out.print(',');
}
}
}
This is basically the same code as yours. However I store the results of the help functions into variables.
A zero is a special case, an exception of the rule that leading zeros are dropped.
Every dth digit (from right) needs to print comma, but not entirely at the right. And not in front. Realized by printing the digit first and then possibly the comma.
The problems I see with your code are the two while loops, twice printing the comma, maybe? And the println with a newline when <= 0.
Test your code, for instance as:
public static void main(String[] args) {
for (int n : new int[] {0, 1, 8, 9, 10, 234,
1_234, 12_345, 123_456, 123_456_789, 1_234_567_890}) {
System.out.printf("%d : ", n);
printWithComma(n, 3);
System.out.println();
}
}
Your code seems overly complicated.
If you've learned about recursion, you can do it like this:
public static void printWithComma(int n, int d) {
printInternal(n, d, 1);
System.out.println();
}
private static void printInternal(int n, int d, int i) {
if (n > 9) {
printInternal(n / 10, d, i + 1);
if (i % d == 0)
System.out.print(',');
}
System.out.print(n % 10);
}
Without recursion:
public static void printWithComma(int n, int d) {
int rev = 0, i = d - 1;
for (int num = n; num > 0 ; num /= 10, i++)
rev = rev * 10 + num % 10;
for (; i > d; rev /= 10, i--) {
System.out.print(rev % 10);
if (i % d == 0)
System.out.print(',');
}
System.out.println(rev);
}
Are you allowed to use the whole Java API?
What about something as simple as using DecimalFormat
double in = 12345678;
DecimalFormat df = new DecimalFormat( ",##" );
System.out.println(df.format(in));
12,34,56,78
Using...
,# = 1 per group
,## = 2 per group
,### = 3 per group
etc...
It took me a bunch of minutes. The following code snippet does the job well (explanation below):
public static void printWithComma(int n, int d) { // n=number, d=commaIndex
final int length = (int) (Math.log10(n) + 1); // number of digits;
for (int i = 1; i < Math.pow(10, length); i*=10) { // loop by digits
double current = Math.log10(i); // current loop
double remains = length - current - 1; // loops remaining
int digit = (int) ((n / Math.pow(10, remains)) % 10); // nth digit
System.out.print(digit); // print it
if (remains % d == 0 && remains > 0) { // add comma if qualified
System.out.print(",");
}
}
}
Using (Math.log10(n) + 1) I find a number of digits in the integer (8 for 12345678).
The for-loop assures the exponents of n series (1, 10, 100, 1000...) needed for further calculations. Using logarithm of base 10 I get the current index of the loop.
To get nth digit is a bit tricky and this formula is based on this answer. Then it is printed out.
Finally, it remains to find a qualified position for the comma (,). If modulo of the current loop index is equal zero, the dth index is reached and the comma can be printed out. Finally the condition remains > 0 assures there will be no comma left at the end of the printed result.
Output:
For 4: 1234,5678
For 3: 12,345,678
For 2: 12,34,56,78
For 1: 1,2,3,4,5,6,7,8
I need to write the following method: accepts two integer parameters and returns an integer. If either integer is not a 4 digit number than the method should return the smaller integer. Otherwise, the method should return a four digit integer made up of the smallest digit in the thousands place, hundreds place, tens place and ones place. We cannot turn the integers into Strings, or use lists, or arrays.
For example biggestLoser(6712,1234) returns 1212
For example biggestLoser(19,8918) returns 19
Here's how I've started to write it:
public static int biggestLoser(int a, int b){
if(a<9999 || b<9999){
if(a<b)
return a;
else if(b<a)
return b;
}
int at=a/1000;
int ah=a%1000/100;
int an=a%100/10;
int ae=a%10;
int bt=b/1000;
int bh=b%1000/100;
int bn=b%100/10;
int be=a%10;
if(at<bt && ah<bh && an<bn && ae<be)
return at*1000+ah*100+an*10+ae;
else if(at<bt && ah<bh && an<bn && be<ae)
return at*1000+ah*100+an*10+be;
else if(at<bt&& ah<bh && bn<an && ae<be)
else return at*1000+ah*100+bn*10+ae;
However, it looks like I'm going to have to write way too many if statements, is there a shorter way to write the code?
public static int biggestLoser(int a, int b) {
if (a < 1000 || a >= 10000 || b < 1000 || b >= 10000) {
return Math.min(a, b);
} else {
// both a and b are four digits
int result = 0 ;
int multiplier = 1 ;
for (int digit = 0; digit < 4; digit++) {
int nextDigit = Math.min(a % 10, b % 10);
result = result + nextDigit * multiplier ;
multiplier = multiplier * 10 ;
a = a / 10 ;
b = b / 10 ;
}
return result ;
}
}
How does this work? a % 10 is the remainder when a is divided by 10: in other words it is the least significant digit of a (the "ones place").
a = a / 10 performs integer division, so it divides a by 10 and ignores any fraction. So 1234 becomes 123, and on the next iteration 123 becomes 12, etc. In other words, it discards the "ones place".
So the first time through the loop, you look at the "ones" from a and b, find the smallest one, and add it to result. Then you drop the "ones" from both a and b. So what used to be the "tens" are now the "ones". The second time through the loop, you get the smallest "ones" again: but this was originally the smallest "tens". You want to add that to result, but you need to multiply by 10. This is the multiplier: each time through the loop the multiplier is multiplied by 10. So each time, you get the smallest "ones", multiply by the correct thing, add to the result, and then drop the "ones" from a and b.
Just for fun, here's an implementation that needs only one statement (and works if you replace "four digits" with any positive number of digits). You can ask your instructor to explain it ;).
public static final int NUM_DIGITS = 4 ;
public static final int MAX = (int) Math.pow(10, NUM_DIGITS) ;
public static final int MIN = MAX / 10 ;
public static int biggestLoser(int a, int b) {
return (a < MIN || a >= MAX || b < MIN || b >= MAX) ? Math.min(a, b) :
IntStream.iterate(1, multiplier -> multiplier * 10).limit(NUM_DIGITS)
.map(multiplier -> Math.min((a / multiplier) % 10, (b / multiplier) % 10) * multiplier )
.sum();
}
maybe it is stupid but try to take advantage of String ( .charAt(int index) )and Integer ( .parseInt( String value ) ) methods , maybe this example help you :
int x=145;
int y=826;
//to know which number have the biggest tens
String a=x+"";
String b=y+"";
if(Integer.parseInt(a.charAt(1)+"")>Integer.parseInt(b.charAt(1)+""))
{
System.out.println("The number which have the biggest tens is "+a);
}
else
{
System.out.println("The number which have the biggest tens is "+b);
}
Using String and StringBuilder
public class Test
{
public static void main(String []args)
{
System.out.println(biggestLooser(6712,1234));
}
public static int biggestLooser(int _a, int _b)
{
String a = String.valueOf(_a);
String b = String.valueOf(_b);
StringBuilder c = new StringBuilder();
if(a.length() < b.length()) return Integer.parseInt(a);
else if(b.length() < a.length()) return Integer.parseInt(b);
else if(a.length() >= 4 && b.length() >= 4)
{
for(int i = 4; i > 0; i--)
{
char ch = '\0';
if(a.charAt(a.length() - i) < b.charAt(b.length() - i))
ch = a.charAt(a.length() - i);
else ch = b.charAt(b.length() - i);
c.append(ch);
}
return Integer.parseInt(c.toString());
}
else return -1;
}
}
//ouput: 1212
Here is the simple answer
public static int biggestLoser(int a, int b) {
if (a < 1000 || b < 1000) {
if (a < b)
return a;
else
return b;
}
int val = 0;
ArrayList<Integer> data1 = new ArrayList<Integer>();
while (a > 0) {
data1.add(a % 10);
a /= 10;
}
Collections.reverse(data1);
ArrayList<Integer> data2 = new ArrayList<Integer>();
while (b > 0) {
data2.add(b % 10);
b /= 10;
}
Collections.reverse(data2);
val = ((data1.get(0) < data2.get(0)) ? data1.get(0) : data2.get(0))
* 1000
+ ((data1.get(1) < data2.get(1)) ? data1.get(1) : data2.get(1))
* 100
+ ((data1.get(2) < data2.get(2)) ? data1.get(2) : data2.get(2))
* 10
+ ((data1.get(3) < data2.get(3)) ? data1.get(3) : data2.get(3));
return val;
}
Given a sequence which contains only various amounts of the numbers 1, 2, 3, and 4 (examples: 13244, 4442, etc), I want to count all its permutations such that no two adjacent numbers are the same. I believe it is O(N! * N) and want to know if there is a better one out there. Anyone have any ideas?
class Ideone
{
static int permutationCount++;
public static void main(String[] args) {
String str = "442213";
permutation("", str);
System.out.println(permutationCount);
}
private static void permutation(String prefix, String str) {
int n = str.length();
if (n == 0){
boolean bad = false;
//Check whether there are repeating adjacent characters
for(int i = 0; i < prefix.length()-1; i++){
if(prefix.charAt(i)==prefix.charAt(i+1))
bad = true;
}
if(!bad){
permutationCount++;
}
}
else {
//Recurse through permutations
for (int i = 0; i < n; i++)
permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n));
}
}
}
I understand your question like this: Given a string containing only numbers 1,2,3,4 - how many permutations of these characters exist that when you put them into string again there won't be any same adjecent numbers.
I would suggest this approach:
L - length of your string
n1 - how many times is 1 repeated, n2 - how many times is 2 repeated etc.
P - number of all possible permutations
P = L! / (n1!*n2!*n3!*n4!)
C - number of all solutions fitting your constraint
C = P - start with all permutations
substract all permutations which have 11 in it (just take 11 as one number)
C = C - (L - 1)! / ((n1 - 1)! * n2! * n3! * n4!)
... do the same for 22 ...
add all permutations which have both 11 and 22 in it (because we have substracted them twice, so you need to add them)
C = C + (L - 2)! / ((n1 - 1)! * (n2 - 1)! * n3! * n4!)
... repeat previous steps for 33 and 44 ...
If you just want to calculate how many permutations match your constraint, it is not required to spell each of them out.
If I get your question right, your input string has 4 distinct input characters 1,2,3,4 and you want to know how many permutations of this are possible?
Then you should use some maths, namely n! / (n-r)!, where n is the number of elements to choose from (4 in this case) and r is the number of positions you want to fill (also 4).
Your example would have 4! / (4-4)! = 24 permutations possible:
{1,2,3,4} {1,2,4,3} {1,3,2,4} {1,3,4,2} {1,4,2,3} {1,4,3,2}
{2,1,3,4} {2,1,4,3} {2,3,1,4} {2,3,4,1} {2,4,1,3} {2,4,3,1}
{3,1,2,4} {3,1,4,2} {3,2,1,4} {3,2,4,1} {3,4,1,2} {3,4,2,1}
{4,1,2,3} {4,1,3,2} {4,2,1,3} {4,2,3,1} {4,3,1,2} {4,3,2,1}
In a nutshell, for length n with n distinct values the count of permutations is n!:
1 -> 1
2 -> 2
3 -> 6
4 -> 24
5 -> 120
...
Edit: After your edits and comments it is clear that I misunderstood the question.
If you just want to check to see if there are no matching adjacent numbers, then you can use a simple loop. This will be O(n) complexity.
public static void main(String[] args) {
String str = "442213";
System.out.println(permutation(str));
}
public static int permutation(String str) {
int permutationCount = 0;
if (str.length() > 1) {
for (int i = 0; i < str.length() - 1; i++) {
if (str.charAt(i) != str.charAt(i + 1)) {
permutationCount++;
}
}
}
return permutationCount;
}
If you wanted to stick with recursion, you could do something like this:
public static void main(String[] args) {
String str = "442213";
System.out.println(permutation(str, 0));
}
public static int permutation(String str, int currentIndex) {
int permutationCount = 0;
if (str == null || currentIndex + 1 >= str.length()) {
return permutationCount;
}
if (str.charAt(currentIndex) != str.charAt(currentIndex + 1)) {
permutationCount = 1;
}
return permutationCount + permutation(str, currentIndex + 1);
}
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int value = 0;
for(int i = 100;i <=999;i++)
{
for(int j = i;j <=999;j++)
{
int value1 = i * j;
StringBuilder sb1 = new StringBuilder(""+value1);
String sb2 = ""+value1;
sb1.reverse();
if(sb2.equals(sb1.toString()) && value<value1) {
value = value1;
}
}
}
System.out.println(value);
}
}
This is the code that I wrote in Java... Is there any efficient way other than this.. And can we optimize this code more??
We suppose the largest such palindrome will have six digits rather than five, because 143*777 = 111111 is a palindrome.
As noted elsewhere, a 6-digit base-10 palindrome abccba is a multiple of 11. This is true because a*100001 + b*010010 + c*001100 is equal to 11*a*9091 + 11*b*910 + 11*c*100. So, in our inner loop we can decrease n by steps of 11 if m is not a multiple of 11.
We are trying to find the largest palindrome under a million that is a product of two 3-digit numbers. To find a large result, we try large divisors first:
We step m downwards from 999, by 1's;
Run n down from 999 by 1's (if 11 divides m, or 9% of the time) or from 990 by 11's (if 11 doesn't divide m, or 91% of the time).
We keep track of the largest palindrome found so far in variable q. Suppose q = r·s with r <= s. We usually have m < r <= s. We require m·n > q or n >= q/m. As larger palindromes are found, the range of n gets more restricted, for two reasons: q gets larger, m gets smaller.
The inner loop of attached program executes only 506 times, vs the ~ 810000 times the naive program used.
#include <stdlib.h>
#include <stdio.h>
int main(void) {
enum { A=100000, B=10000, C=1000, c=100, b=10, a=1, T=10 };
int m, n, p, q=111111, r=143, s=777;
int nDel, nLo, nHi, inner=0, n11=(999/11)*11;
for (m=999; m>99; --m) {
nHi = n11; nDel = 11;
if (m%11==0) {
nHi = 999; nDel = 1;
}
nLo = q/m-1;
if (nLo < m) nLo = m-1;
for (n=nHi; n>nLo; n -= nDel) {
++inner;
// Check if p = product is a palindrome
p = m * n;
if (p%T==p/A && (p/B)%T==(p/b)%T && (p/C)%T==(p/c)%T) {
q=p; r=m; s=n;
printf ("%d at %d * %d\n", q, r, s);
break; // We're done with this value of m
}
}
}
printf ("Final result: %d at %d * %d inner=%d\n", q, r, s, inner);
return 0;
}
Note, the program is in C but same techniques will work in Java.
What I would do:
Start at 999, working my way backwards to 998, 997, etc
Create the palindrome for my current number.
Determine the prime factorization of this number (not all that expensive if you have a pre-generated list of primes.
Work through this prime factorization list to determine if I can use a combination of the factors to make 2 3 digit numbers.
Some code:
int[] primes = new int[] {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,,127,131,137,139,149,151,157,163,167,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,
947,953,967,971,977,983,991,997};
for(int i = 999; i >= 100; i--) {
String palstr = String.valueOf(i) + (new StringBuilder().append(i).reverse());
int pal = Integer.parseInt(pal);
int[] factors = new int[20]; // cannot have more than 20 factors
int remainder = pal;
int facpos = 0;
primeloop:
for(int p = 0; p < primes.length; i++) {
while(remainder % p == 0) {
factors[facpos++] = p;
remainder /= p;
if(remainder < p) break primeloop;
}
}
// now to do the combinations here
}
We can translate the task into the language of mathematics.
For a short start, we use characters as digits:
abc * xyz = n
abc is a 3-digit number, and we deconstruct it as 100*a+10*b+c
xyz is a 3-digit number, and we deconstruct it as 100*x+10*y+z
Now we have two mathematical expressions, and can define a,b,c,x,y,z as € of {0..9}.
It is more precise to define a and x as of element from {1..9}, not {0..9}, because 097 isn't really a 3-digit number, is it?
Ok.
If we want to produce a big number, we should try to reach a 9......-Number, and since it shall be palindromic, it has to be of the pattern 9....9. If the last digit is a 9, then from
(100*a + 10*b + c) * (100*x + 10*y + z)
follows that z*c has to lead to a number, ending in digit 9 - all other calculations don't infect the last digit.
So c and z have to be from (1,3,7,9) because (1*9=9, 9*1=9, 3*3=9, 7*7=49).
Now some code (Scala):
val n = (0 to 9)
val m = n.tail // 1 to 9
val niners = Seq (1, 3, 7, 9)
val highs = for (a <- m;
b <- n;
c <- niners;
x <- m;
y <- n;
z <- niners) yield ((100*a + 10*b + c) * (100*x + 10*y + z))
Then I would sort them by size, and starting with the biggest one, test them for being palindromic. So I would omit to test small numbers for being palindromic, because that might not be so cheap.
For aesthetic reasons, I wouldn't take a (toString.reverse == toString) approach, but a recursive divide and modulo solution, but on todays machines, it doesn't make much difference, does it?
// Make a list of digits from a number:
def digitize (z: Int, nums : List[Int] = Nil) : List[Int] =
if (z == 0) nums else digitize (z/10, z%10 :: nums)
/* for 342243, test 3...==...3 and then 4224.
Fails early for 123329 */
def palindromic (nums : List[Int]) : Boolean = nums match {
case Nil => true
case x :: Nil => true
case x :: y :: Nil => x == y
case x :: xs => x == xs.last && palindromic (xs.init) }
def palindrom (z: Int) = palindromic (digitize (z))
For serious performance considerations, I would test it against a toString/reverse/equals approach. Maybe it is worse. It shall fail early, but division and modulo aren't known to be the fastest operations, and I use them to make a List from the Int. It would work for BigInt or Long with few redeclarations, and works nice with Java; could be implemented in Java but look different there.
Okay, putting the things together:
highs.filter (_ > 900000) .sortWith (_ > _) find (palindrom)
res45: Option[Int] = Some(906609)
There where 835 numbers left > 900000, and it returns pretty fast, but I guess even more brute forcing isn't much slower.
Maybe there is a much more clever way to construct the highest palindrom, instead of searching for it.
One problem is: I didn't knew before, that there is a solution > 900000.
A very different approach would be, to produce big palindromes, and deconstruct their factors.
public class Pin
{
public static boolean isPalin(int num)
{
char[] val = (""+num).toCharArray();
for(int i=0;i<val.length;i++)
{
if(val[i] != val[val.length - i - 1])
{
return false;
}
}
return true;
}
public static void main(String[] args)
{
for(int i=999;i>100;i--)
for(int j=999;j>100;j--)
{
int mul = j*i;
if(isPalin(mul))
{
System.out.printf("%d * %d = %d",i,j,mul);
return;
}
}
}
}
package ex;
public class Main {
public static void main(String[] args) {
int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, flag = 0;
for (i = 999; i >= 100; i--) {
for (j = i; j >= 100; j--) {
k = i * j;
// System.out.println(k);
m = 0;
n = k;
while (n > 0) {
l = n % 10;
m = m * 10 + l;
n = n / 10;
}
if (m == k) {
System.out.println("pal " + k + " of " + i + " and" + j);
flag = 1;
break;
}
}
if (flag == 1) {
// System.out.println(k);
break;
}
}
}
}
A slightly different approach that can easily calculate the largest palindromic number made from the product of up to two 6-digit numbers.
The first part is to create a generator of palindrome numbers. So there is no need to check if a number is palindromic, the second part is a simple loop.
#include <memory>
#include <iostream>
#include <cmath>
using namespace std;
template <int N>
class PalindromeGenerator {
unique_ptr <int []> m_data;
bool m_hasnext;
public :
PalindromeGenerator():m_data(new int[N])
{
for(auto i=0;i<N;i++)
m_data[i]=9;
m_hasnext=true;
}
bool hasNext() const {return m_hasnext;}
long long int getnext()
{
long long int v=0;
long long int b=1;
for(int i=0;i<N;i++){
v+=m_data[i]*b;
b*=10;
}
for(int i=N-1;i>=0;i--){
v+=m_data[i]*b;
b*=10;
}
auto i=N-1;
while (i>=0)
{
if(m_data[i]>=1) {
m_data[i]--;
return v;
}
else
{
m_data[i]=9;
i--;
}
}
m_hasnext=false;
return v;
}
};
template<int N>
void findmaxPalindrome()
{
PalindromeGenerator<N> gen;
decltype(gen.getnext()) minv=static_cast<decltype(gen.getnext())> (pow(10,N-1));
decltype(gen.getnext()) maxv=static_cast<decltype(gen.getnext())> (pow(10,N)-1);
decltype(gen.getnext()) start=11*(maxv/11);
while(gen.hasNext())
{
auto v=gen.getnext();
for (decltype(gen.getnext()) i=start;i>minv;i-=11)
{
if (v%i==0)
{
auto r=v/i;
if (r>minv && r<maxv ){
cout<<"done:"<<v<<" "<<i<< "," <<r <<endl;
return ;
}
}
}
}
return ;
}
int main(int argc, char* argv[])
{
findmaxPalindrome<6>();
return 0;
}
You can use the fact that 11 is a multiple of the palindrome to cut down on the search space. We can get this since we can assume the palindrome will be 6 digits and >= 111111.
e.g. ( from projecteuler ;) )
P= xyzzyx = 100000x + 10000y + 1000z + 100z + 10y +x
P=100001x+10010y+1100z
P=11(9091x+910y+100z)
Check if i mod 11 != 0, then the j loop can be subtracted by 11 (starting at 990) since at least one of the two must be divisible by 11.
You can try the following which prints
999 * 979 * 989 = 967262769
largest palindrome= 967262769 took 0.015
public static void main(String... args) throws IOException, ParseException {
long start = System.nanoTime();
int largestPalindrome = 0;
for (int i = 999; i > 100; i--) {
LOOP:
for (int j = i; j > 100; j--) {
for (int k = j; k > 100; k++) {
int n = i * j * k;
if (n < largestPalindrome) continue LOOP;
if (isPalindrome(n)) {
System.out.println(i + " * " + j + " * " + k + " = " + n);
largestPalindrome = n;
}
}
}
}
long time = System.nanoTime() - start;
System.out.printf("largest palindrome= %d took %.3f seconds%n", largestPalindrome, time / 1e9);
}
private static boolean isPalindrome(int n) {
if (n >= 100 * 1000 * 1000) {
// 9 digits
return n % 10 == n / (100 * 1000 * 1000)
&& (n / 10 % 10) == (n / (10 * 1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (1000 * 1000) % 10)
&& (n / 1000 % 10) == (n / (100 * 1000) % 10);
} else if (n >= 10 * 1000 * 1000) {
// 8 digits
return n % 10 == n / (10 * 1000 * 1000)
&& (n / 10 % 10) == (n / (1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (100 * 1000) % 10)
&& (n / 1000 % 10) == (n / (10 * 1000) % 10);
} else if (n >= 1000 * 1000) {
// 7 digits
return n % 10 == n / (1000 * 1000)
&& (n / 10 % 10) == (n / (100 * 1000) % 10)
&& (n / 100 % 10) == (n / (10 * 1000) % 10);
} else throw new AssertionError();
}
i did this my way , but m not sure if this is the most efficient way of doing this .
package problems;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P_4 {
/**
* #param args
* #throws IOException
*/
static int[] arry = new int[6];
static int[] arry2 = new int[6];
public static boolean chk()
{
for(int a=0;a<arry.length;a++)
if(arry[a]!=arry2[a])
return false;
return true;
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int temp,z,i;
for(int x=999;x>100;x--)
for(int y=999;y>100;y--)
{
i=0;
z=x*y;
while(z>0)
{
temp=z%10;
z=z/10;
arry[i]=temp;
i++;
}
for(int k = arry.length;k>0;k--)
arry2[arry.length- k]=arry[k-1];
if(chk())
{
System.out.print("pelindrome = ");
for(int l=0;l<arry2.length;l++)
System.out.print(arry2[l]);
System.out.println(x);
System.out.println(y);
}
}
}
}
This is code in C, a little bit long, but gets the job done.:)
#include <stdio.h>
#include <stdlib.h>
/*
A palindromic number reads the same both ways. The largest palindrome made from the product of two
2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.*/
int palndr(int b)
{
int *x,*y,i=0,j=0,br=0;
int n;
n=b;
while(b!=0)
{
br++;
b/=10;
}
x=(int *)malloc(br*sizeof(int));
y=(int *)malloc(br*sizeof(int));
int br1=br;
while(n!=0)
{
x[i++]=y[--br]=n%10;
n/=10;
}
int ind = 1;
for(i=0;i<br1;i++)
if(x[i]!=y[i])
ind=0;
free(x);
free(y);
return ind;
}
int main()
{
int i,cek,cekmax=1;
int j;
for(i=100;i<=999;i++)
{
for(j=i;j<=999;j++)
{
cek=i*j;
if(palndr(cek))
{
if(pp>cekmax)
cekmax=cek;
}
}
}
printf("The largest palindrome is: %d\n\a",cekmax);
}
You can actually do it with Python, it's easy just take a look:
actualProduct = 0
highestPalindrome = 0
# Setting the numbers. In case it's two digit 10 and 99, in case is three digit 100 and 999, etc.
num1 = 100
num2 = 999
def isPalindrome(number):
number = str(number)
reversed = number[::-1]
if number==reversed:
return True
else:
return False
a = 0
b = 0
for i in range(num1,num2+1):
for j in range(num1,num2+1):
actualProduct = i * j
if (isPalindrome(actualProduct) and (highestPalindrome < actualProduct)):
highestPalindrome = actualProduct
a = i
b = j
print "Largest palindrome made from the product of two %d-digit numbers is [ %d ] made of %d * %d" % (len(str(num1)), highestPalindrome, a, b)
Since we are not cycling down both iterators (num1 and num2) at the same time, the first palindrome number we find will be the largest. We don’t need to test to see if the palindrome we found is the largest. This significantly reduces the time it takes to calculate.
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int limit = 99;
int max = 999;
int num1 = max, num2, prod;
while(num1 > limit)
{
num2 = num1;
while(num2 > limit)
{
total = num1 * num2;
StringBuilder sb1 = new StringBuilder(""+prod);
String sb2 = ""+prod;
sb1.reverse();
if( sb2.equals(sb1.toString()) ) { //optimized here
//print and exit
}
num2--;
}
num1--;
}
}//end of main
}//end of class PalindromeThreeDigits
I tried the solution by Tobin joy and vickyhacks and both of them produce the result 580085 which is wrong here is my solution, though very clumsy:
import java.util.*;
class ProjEu4
{
public static void main(String [] args) throws Exception
{
int n=997;
ArrayList<Integer> al=new ArrayList<Integer>();
outerloop:
while(n>100){
int k=reverse(n);
int fin=n*1000+k;
al=findfactors(fin);
if(al.size()>=2)
{
for(int i=0;i<al.size();i++)
{
if(al.contains(fin/al.get(i))){
System.out.println(fin+" factors are:"+al.get(i)+","+fin/al.get(i));
break outerloop;}
}
}
n--;
}
}
private static ArrayList<Integer> findfactors(int fin)
{
ArrayList<Integer> al=new ArrayList<Integer>();
for(int i=100;i<=999;i++)
{
if(fin%i==0)
al.add(i);
}
return al;
}
private static int reverse(int number)
{
int reverse = 0;
while(number != 0){
reverse = (reverse*10)+(number%10);
number = number/10;
}
return reverse;
}
}
Most probably it is replication of one of the other solution but it looks simple owing to pythonified code ,even it is a bit brute-force.
def largest_palindrome():
largest_palindrome = 0;
for i in reversed(range(1,1000,1)):
for j in reversed(range(1, i+1, 1)):
num = i*j
if check_palindrome(str(num)) and num > largest_palindrome :
largest_palindrome = num
print "largest palindrome ", largest_palindrome
def check_palindrome(term):
rev_term = term[::-1]
return rev_term == term
What about : in python
>>> for i in range((999*999),(100*100), -1):
... if str(i) == str(i)[::-1]:
... print i
... break
...
997799
>>>
I believe there is a simpler approach: Examine palindromes descending from the largest product of two three digit numbers, selecting the first palindrome with two three digit factors.
Here is the Ruby code:
require './palindrome_range'
require './prime'
def get_3_digit_factors(n)
prime_factors = Prime.factors(n)
rf = [prime_factors.pop]
rf << prime_factors.shift while rf.inject(:*) < 100 || prime_factors.inject(:*) > 999
lf = prime_factors.inject(:*)
rf = rf.inject(:*)
lf < 100 || lf > 999 || rf < 100 || rf > 999 ? [] : [lf, rf]
end
def has_3_digit_factors(n)
return !get_3_digit_factors(n).empty?
end
pr = PalindromeRange.new(0, 999 * 999)
n = pr.downto.find {|n| has_3_digit_factors(n)}
puts "Found #{n} - Factors #{get_3_digit_factors(n).inspect}, #{Prime.factors(n).inspect}"
prime.rb:
class Prime
class<<self
# Collect all prime factors
# -- Primes greater than 3 follow the form of (6n +/- 1)
# Being of the form 6n +/- 1 does not mean it is prime, but all primes have that form
# See http://primes.utm.edu/notes/faq/six.html
# -- The algorithm works because, while it will attempt non-prime values (e.g., (6 *4) + 1 == 25),
# they will fail since the earlier repeated division (e.g., by 5) means the non-prime will fail.
# Put another way, after repeatedly dividing by a known prime, the remainder is itself a prime
# factor or a multiple of a prime factor not yet tried (e.g., greater than 5).
def factors(n)
square_root = Math.sqrt(n).ceil
factors = []
while n % 2 == 0
factors << 2
n /= 2
end
while n % 3 == 0
factors << 3
n /= 3
end
i = 6
while i < square_root
[(i - 1), (i + 1)].each do |f|
while n % f == 0
factors << f
n /= f
end
end
i += 6
end
factors << n unless n == 1
factors
end
end
end
palindrome_range.rb:
class PalindromeRange
FIXNUM_MAX = (2**(0.size * 8 -2) -1)
def initialize(min = 0, max = FIXNUM_MAX)
#min = min
#max = max
end
def downto
return enum_for(:downto) unless block_given?
n = #max
while n >= #min
yield n if is_palindrome(n)
n -= 1
end
nil
end
def each
return upto
end
def upto
return enum_for(:downto) unless block_given?
n = #min
while n <= #max
yield n if is_palindrome(n)
n += 1
end
nil
end
private
def is_palindrome(n)
s = n.to_s
i = 0
j = s.length - 1
while i <= j
break if s[i] != s[j]
i += 1
j -= 1
end
i > j
end
end
public class ProjectEuler4 {
public static void main(String[] args) {
int x = 999; // largest 3-digit number
int largestProduct = 0;
for(int y=x; y>99; y--){
int product = x*y;
if(isPalindormic(x*y)){
if(product>largestProduct){
largestProduct = product;
System.out.println("3-digit numbers product palindormic number : " + x + " * " + y + " : " + product);
}
}
if(y==100 || product < largestProduct){y=x;x--;}
}
}
public static boolean isPalindormic(int n){
int palindormic = n;
int reverse = 0;
while(n>9){
reverse = (reverse*10) + n%10;
n=n/10;
}
reverse = (reverse*10) + n;
return (reverse == palindormic);
}
}